initial commit / add all books

10 files changed, 245 insertions, 0 deletions

diff --git a/1910/CH1/EX1.1/Chapter11.sce b/1910/CH1/EX1.1/Chapter11.sce
new file mode 100755
index 000000000..85e657cd2
--- /dev/null
+++ b/1910/CH1/EX1.1/Chapter11.sce
@@ -0,0 +1,17 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 1")

+//The temprature of two faces of the slabs are T1=40°C & T2=20°C 

+//The thickness of the slab(L) is 80mm or .08m

+//The thermal conductivity(k)of the material is .20 W/(m*K)

+T1=40;

+T2=20;

+L=.08;

+k=.20;

+//The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L

+disp ("The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m^2 ")

+q=k*(T1-T2)/L

diff --git a/1910/CH1/EX1.10/Chapter110.sce b/1910/CH1/EX1.10/Chapter110.sce
new file mode 100755
index 000000000..ff7a2511e
--- /dev/null
+++ b/1910/CH1/EX1.10/Chapter110.sce
@@ -0,0 +1,49 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 10")

+//The spacecraft panel has thickness(L)=.01 m

+//The spacecraft has inner temprature (Ti)=298 K

+//The spacecraft has outer temprature(T2)

+//The panel is exposed to deep space where temprature(To)= 0K

+//The material has Thermal conductivity(k)= 5.0 W/(m*K)

+//The emissivity(emi)=0.8

+//The inner surface of the panel is exposed to airflow resulting in an average heat transfer coefficient(hbri)=70 W/(m^2*K)

+L=.01;

+Ti=298;

+To=0;

+k=5;

+emi=0.8;

+hbri=70;

+//The stefan Boltzman constant(sigma)= 5.67*10^-8 W/(m^2/K^4)

+sigma=5.67*10^-8;

+//Heat transfer from the outer surface takes place only by radiation is given by  Q/A=emi*sigma*(T2^4-T0^4)in W/m^2=F1

+//heat transfer from the outer surface can also be written as Q/A=(Ti-To)/((1/hbri)+(L/k)+(1/hr))=F2

+//Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)

+//so hr=4.536*10^-8*T2^3

+disp("Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2^4-T0^4)in W/m^2 for different values of tempratures in K")

+disp("heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m^2 at different tempratures in K" )

+disp("The values of temprature that are considered are <298 K")

+for (i=285:292)

+    T2=i,hr=4.536*10^-8*T2^3; F1=emi*sigma*(T2^4-To^4),F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))

+end

+if F1==F2 then T2=i

+    else T2=292.5,hr=4.536*10^-8*T2^3; F1=emi*sigma*(T2^4-To^4),F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))

+end

+disp("Satisfactory solutions for Temprature in K is")

+disp(Temprature = T2)

+disp("Approximate Rate of Heat Transfer in W/m^2 is")

+disp(332)

+

+

+

+

+

+

+

+

+

+

diff --git a/1910/CH1/EX1.11/Chapter111.sce b/1910/CH1/EX1.11/Chapter111.sce
new file mode 100755
index 000000000..f26bc4e83
--- /dev/null
+++ b/1910/CH1/EX1.11/Chapter111.sce
@@ -0,0 +1,27 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 11")

+//The horizontal steel pipe has outer diameter(D)=80 mm or.08 m

+//The pipe is maintained at a temprature(T1)=60°C where the air and wall temprature(T2)=20 °C 

+//The average free convective heat transfer coefficient(hbr)=6.5 W/(m^2/K) b/w the outer surface of the pipe and air

+D=.08;

+T1=60;

+T2=20;

+hbr=6.5;

+//Length(L=1) since per unit length is considered

+L=1;

+//The surface area of pipe is given by A=(%pi*D*L)

+A=(%pi*D*L);

+//The surface emissivity(emi) of steel = 0.8

+//The stefan -Boltzman constant(sigma)= 5.7*10^-8 W/(m^2*K^4)

+sigma=5.67*10^-8;

+emi=.8;

+//The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1^4-T2^4)

+disp("The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1^4-T2^4) in W/m")

+//Let Q/L=F

+F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)^4-(T2+273.15)^4)

+

diff --git a/1910/CH1/EX1.2/Chapter12.sce b/1910/CH1/EX1.2/Chapter12.sce
new file mode 100755
index 000000000..6f7eef26e
--- /dev/null
+++ b/1910/CH1/EX1.2/Chapter12.sce
@@ -0,0 +1,20 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 2")

+//The thermal conductivity(km)of masonry wall is .8 W/(mK)

+//The thermal conductivity(kc)of composite wall is .2 W/(mK)

+//The thickness of composite wall(Lc) is 100 mm or .1 m

+km=.8;

+kc=.2;

+Lc=.1;

+//The thickness of masonry wall(Lm) is to be found. 

+//The steady state heat flow(qm)through masonry wall is km(T1-T2)/L

+// The steady state heat flow(qc)through composite wall is kc(T1-T2)/L

+//As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)

+//The thickness of masonry wall is Lm.

+disp ("The thickness of masonry wall is Lm in m")

+Lm=(km/kc)*(Lc/(0.8))

diff --git a/1910/CH1/EX1.4/Chapter14.sce b/1910/CH1/EX1.4/Chapter14.sce
new file mode 100755
index 000000000..ababfe72d
--- /dev/null
+++ b/1910/CH1/EX1.4/Chapter14.sce
@@ -0,0 +1,16 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 4")

+//The average forced convective heat transfer coefficient(hbr) is 200 W/( m^2 °C)

+//The fluid temprature(Tinf) upstream of the cold surface is 100°C

+//The surface temprature(Ts) is 20°C

+hbr=200;

+Tinf=100;

+Ts=20;

+//The rate of heat transfer per unit area is q

+disp ("The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m^2")

+q=hbr*(Tinf-Ts)

diff --git a/1910/CH1/EX1.5/Chapter15.sce b/1910/CH1/EX1.5/Chapter15.sce
new file mode 100755
index 000000000..9e4910a43
--- /dev/null
+++ b/1910/CH1/EX1.5/Chapter15.sce
@@ -0,0 +1,16 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 5")

+//The average heat transfer coefficient(hbr) is 800 W/(m^2°C)

+//The surface temprature of heat exchanger is 75°C and air temprature is 25°C so deltaT=(75-25)

+//The amount of heat exchanged(Q) is 20 MJ/h

+//The heat exchanger surface area(A) is given by A=Q/(hbr*∆T)

+hbr=800;

+deltaT=(75-25);

+Q=20;

+disp("The heat exchanger surface area(A)in m^2 required for 20 MJ/h of heating is ")

+A = (Q*10^6)/(3600*hbr*deltaT)

diff --git a/1910/CH1/EX1.6/Chapter16.sce b/1910/CH1/EX1.6/Chapter16.sce
new file mode 100755
index 000000000..953c5daaf
--- /dev/null
+++ b/1910/CH1/EX1.6/Chapter16.sce
@@ -0,0 +1,39 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 6")

+//The temprature of the plate(Ts) is 225°C

+//The ambient temprature (Tinf) is 25°C

+//The change in plate temprature with time is dT/dt=-.02K/s

+//The plate area (A)=.1m^2 , mass(m)= 4Kg and specific heat(cp)=2.8KJ/(Kg*K)

+//The average free convective heat coefficient(hbr) is to be found

+Ts=225;

+Tinf=25;

+//|dT/dt|=0.2,because it is modulus function and it converts negative values to positive value.

+//Let |dT/dt|=X

+X=0.02;

+A=.1;

+m=4;

+cp=2.8;

+disp("The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)")

+disp("The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.")

+disp("Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m^2°C)")

+hbr=(m*cp*10^3*X)/(A*(Ts-Tinf))

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

diff --git a/1910/CH1/EX1.7/Chapter17.sce b/1910/CH1/EX1.7/Chapter17.sce
new file mode 100755
index 000000000..9c5862730
--- /dev/null
+++ b/1910/CH1/EX1.7/Chapter17.sce
@@ -0,0 +1,17 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 7")

+//The temprature(T) of brick wall after sunset is 50°C

+//The emissity value(emi)=0.9

+//The radiant heat flux per square meter =E/A Where E is radiant heat energy and A is area of brick wall.

+//The stefan-Boltzman constant(sigma)=5.6697*10^-8 W/(m^2*K^4).

+T=50;

+emi=.9;

+sigma=5.6697*10^-8;

+disp("The heat flux per square meter is given by E/A=emi*sigma*T^4 in W/m^2")

+//Let E/A=F

+F=emi*sigma*(T+273.15)^4

diff --git a/1910/CH1/EX1.8/Chapter18.sce b/1910/CH1/EX1.8/Chapter18.sce
new file mode 100755
index 000000000..5c4c7db64
--- /dev/null
+++ b/1910/CH1/EX1.8/Chapter18.sce
@@ -0,0 +1,15 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 8")

+//The temprature(T) of asphalt pavement = 50°C

+//The stefan-Boltzman constant(sigma)=5.6697*10^-8 W/(m^2*K^4).

+T=50;

+sigma=5.6697*10^-8;

+//The emitted radiant energy per unit surface area is given by (Eb/A)=sigma*T^4

+disp ("The emitted radiant energy per unit surface area is given by Eb/A=sigma*T^4 in W/m^2")

+//Let Eb/A=F

+F=sigma*(50+273.15)^4

diff --git a/1910/CH1/EX1.9/Chapter19.sce b/1910/CH1/EX1.9/Chapter19.sce
new file mode 100755
index 000000000..761767c11
--- /dev/null
+++ b/1910/CH1/EX1.9/Chapter19.sce
@@ -0,0 +1,29 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 9")

+//The Thickness(L) of wall= 150 mm or 0.15 m.

+//The wall on one side is exposed to air at temprature(Ta)= 60°C and on the other side to air at temprature(Tb) = 20°C

+//The average convective heat transfer coefficients are hbr1=40 W/(m^2°C) on the 60°C and hbr2= 10 W/(m^2°C) on 20°C side.

+//The thermal conductivity(k)=.8 W/(m°C)

+L=0.15;

+Ta=60;

+Tb=20;

+hbr1=40;

+hbr2=10;

+k=0.8;

+//Area(A=1 m^2 )since unit surface area is required.

+A=1;

+//The rate of heat transfer per unit surface area of wall is given by (Q/A)=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))

+disp("The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m^2")

+//Let Q/A=F

+F=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))

+//The surface tempratures of wall on 60°C side is T1 and on 20°C side is T2

+disp("The surface tempratures of wall on 60°C side is T1 =Ta-(Q/(A*hbr1)) in °C")

+T1 =Ta-(F/hbr1)

+disp("The surface tempratures of wall on 20°C side is T2 =Tb+(Q/(A*hbr2)) in °C")

+T2 =Tb+(F/hbr2)

+